ME3122E - Tutorial Solution 3

ME3122 Solutions to Tutorial Set 3 1. Using the energy integral equation determine an expression for the heat transfer coefficient by assuming the following velocity and temperature profiles: u  u  constant and (T  TW ) /(T  TW )  y /  t where  t is the thermal boundary layer thickness. Solution: Energy Integral Equation:

d dx Given

Therefore,

T  TW y ,  T  TW t

 uT  T dy  Ty t



0

y 0

T 1  .(T  TW ) y  t

T  TW T  T y  1 1 T  TW T  TW t

Substituting the above and u  u into energy integral equation, we get

u T  TW 

  d  t  y k 1 0   1dy   T  TW  dx    t C p  t  

Integrating u d k 1 ( t )  C p  t 2 dx d ( t2 ) 4k   C p u dx

or

 t2 

4kx ;    C pu t

4kx  C pu

Now:

k

T    h(Tw  T ) y  w

qcoduction = qconvection

h(Tw  T )   k (T  Tw )  4kx h k  C u t p   k

  

1/ 2

1

t 1/ 2

  C p u k     4x 

1

2. Glycerine at 30C flows past a 30cm square flat plate at a velocity of 1.5 m/s. The drag force is measured as 10.98 N (for both sides of the plate). Calculate the heat transfer coefficient for such a flow system. At 30C, properties of glycerine are:   1 2 5 8 k g / m 3 ; C p  2 .4 5 5 k J / k g K ; v  0 .0 0 0 5 0 m 2 / s ;

k  0 .2 6 8 W / m K ;

Pr  5380

Solution:

Reynolds Analogy:

St x Pr 2/3 =

Cf x 2

;

where

St x 

hx  C p u

For one side of the plate: F 10.98   61N / m2 2 A 2  0.32  2  61 C fx  1 w 2   0.0431 u 1258  1.522  2

w 

Substituting:

St x 

0.5  0.0431  7.015  105 2/3 5380

Therefore: hx  7.015  10 5   C p u  7.015  10 5  1258  2445  1.5  323.68W / m 2 K

3. Atmospheric air at 25C flows over a plate at a velocity of 60 m/s. The plate of width 1m and length 0.75m is maintained at a uniform temperature of 230C by independentlycontrolled, electrical strip heaters, each of which is 50mm long in the direction of the airflow. a) Determine at which heater does the flow undergo a transition from laminar to turbulent flow. b) Determine at which heater is the heat input a maximum c) Determine the value of this heat input d) Determine the heat input for the first heater e) Determine the heat input for the first three heaters f) Determine the heat input for the entire plate

2

Solution:

Air u∞ = 60 m/s T∞ = 25C x

1

2

3

Heaters x=L

xc

hx

Turbulent b.l hx ~ x-0.2

Laminar b.l hx ~ x-0.5

xc

x

Given:

T  25 C ; Tw  230 C ; w  1m; L  0.75m u  60m / s; Heater length  0.05m T f  (Tw  T ) / 2  127.5 C  400 K Rogers and Mayhew tables (air ) gives c p  1.0135kJ / kgK ;   2.286 105 kg / m.s; k  3.365 102 W / mK ;   0.8824kg / m3 ; Pr  0.688 a)

Rec  5 105  u xc /  . Substituting gives xc =0.216m

Since each heater is 0.05m, therefore transition to turbulent flow occurs at heater No.5 b) hx at 6th heater (turbulent flow) is highest, because at 5th heater hx is the average of the laminar and turbulent flow values ---as shown in graph.

3

c)

To find hx over the 6th heater, find the hx at its midpoint, i.e. at x=0.275m.

hx .x  0.0296 Re x 0.8 Pr1/ 3 k for turbulent b.l flow

local Nu x 

0.8

 u x  k hx   0.0296    0.6881/ 3 @ x  0.275 x    hx  140.6W / m 2 K Q  hx l.1(230  25)  1441W

d) Over first heater --- laminar flow h .x average Nu x  x  0.664 Re x1/ 2 Pr1/ 3 k for x  L  0.05 for laminar b.l. flow

Substituting : hx  134.1W / m2 K Q  134.1 (0.05 1)(205)  1375W e) Over first 3 heaters--- laminar flow h .x Nu L  L  0.664 Re x1/ 2 Pr1/ 3  2 Nu x k for x  L  3  0.05  0.15m :

x L

hL  77.5W / m 2 K Q  77.5  (0.15  1)(205)  2383W f) Over the entire plate flow is laminar-turbulent h L Average Nu L  L.  (0.037 Re L 0.8  871) Pr1/ 3 k for (mixed) laminar-turbulent b.l. flow k hL  (0.037 Re L 0.8  871) Pr1/ 3 ; L for x  L  0.75m : Re L 0.8  98139

; hL  109.3

Q  hL .L.(230  25)  16.81kW Note: for Part c, to be precise, apply Nu L equation for L= 5 x 0.05 and for L= 6 x 0.05 to get hL for each case. Determine Q1 for L = 0 to L = 0.25, and Q2 for L = 0 to L = 0.30. Then get Q2 – Q1. The difference between this and our more convenient method is small.

4

4. A light breeze at 4.47 m/s blows across a metal building. The height of the building is 3.7m and the width is 6.1m. A net energy flux of 347 W/m2 from the sun is absorbed in the wall and subsequently dissipated to the surroundings by convection. Assuming that the air is at 27C and 1 atmosphere, estimate the average temperature that the wall will attain under equilibrium conditions.

Solution: Air u∞= 4.47m/s T∞= 27C, x Tw=?

qw” =347 W/m2

T  27 C  300 K ; L  6.1m ( area  6.1  3.70) u  4.47 m / s; Constant q  347W / m 2 wall temperature Tw and film temperature T f  (Tw  T ) / 2 are unknown and to be determined.

Approach: Approximation / assumption is reqd . i ) First assume T f  T  300 K or any reasonable value Properties of air at 300K:

  1.846 105 kg / m.s; k  2.264 102 W / mK ;   1.177kg / m3 ; Pr  0.707  u L 1.177  4.47  6.1 Re L     1.74  106  5  105 5 1.846  10  laminar - turbulent flow over flat wall

5

hL. L  (0.037 Re L 0.8  871) Pr1/ 3 k above expression is valid for isothermal wall but is assumed average Nu L 

also valid for constant heat flux wall  hL  10.68W / m 2 K q. A  347  ( A)  hL A.(Tw  27) Tw  59.5 C ii ) To improve accuracy, we may repeat calculations with T f  (Tw  T ) / 2  (27  59.5) / 2  43.25 C  316.25K

5. Engine oil at the rate of 0.02 kg/s flows through a 3-mm diameter tube, 30m long. The oil has an inlet temperature of 60C, while the tube wall temperature is maintained at 100C by a stream condensing on its outer surface. a) Estimate the average heat transfer coefficient for the internal flow of oil. b) Determine the outlet temperature of the oil. Take the properties of engine oil to be: cp = 2131 J/kgK,  = 852 kg/m3,  = 0.375x10-4 m2/s, k = 0.138 W/mK, Pr = 490.

Solution: Engine oil =0.02 kg/s 1

Tb1=600C

Tw=100C=const d=0.003m L=30m

Tw=const 2

Tb2=?

Tb2=? Tb1=60C

Tube flow, constant Tw case (neglect entrance effects) Properties of engine oil are given as : cp = 2131 J/kgK,  = 852 kg/m3,  = 0.375x10-4 m2/s, k = 0.138 W/mK, Pr = 490.

6

Re 

 u d 4m    d

4  0.02  107  256.6